Question

1. a solid substance has a mass of 250.00 g. it is cooled by 25.00°c and loses 4.937 kj of heat. what is the specific heat capacity of the substance? identify the substance using the values in table 5.1.

Explanation:

Step1: Recall the heat - loss formula

The formula for heat loss is \(Q = mc\Delta T\), where \(Q\) is the heat lost, \(m\) is the mass of the substance, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
We need to solve for \(c\), so we can rewrite the formula as \(c=\frac{Q}{m\Delta T}\).

Step2: Convert the units

First, convert the mass \(m = 250.00\space g=0.25000\space kg\) (since \(1\space kg = 1000\space g\)) and \(Q=- 4.937\space kJ=-4937\space J\) (negative because heat is lost), \(\Delta T=- 25.00^{\circ}C\) (negative because the substance is cooled).

Step3: Substitute the values into the formula

Substitute \(m = 0.25000\space kg\), \(Q=-4937\space J\), and \(\Delta T=- 25.00^{\circ}C\) into \(c=\frac{Q}{m\Delta T}\).

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Answer:

The specific heat capacity of the substance is \(790\space J/(kg\cdot^{\circ}C)\). Looking at common specific - heat - capacity values (assuming a table similar to standard ones), this value is close to the specific heat capacity of iron (\(c = 450\space J/(kg\cdot^{\circ}C)\) is wrong, but if we assume some approximation errors in measurement or a non - standard table), if we consider a more accurate match, if the table has values like for granite (\(c\approx 790\space J/(kg\cdot^{\circ}C)\)), then the substance could be granite.